Mark Underwood wrote:

> <jens.k.a@...> wrote:

>> The primes are x+/-n, for n = 2, 22, 26, 38, 46, 58, 62, 82.

>> The first occurrence is at x = 9894203506406653455.

>

> Interesting. Jens I honestly don't know how it is you can check all

> the way up to that huge number!

I used an unpublished tuple search program which has found hundreds

of tuple patterns since 2003.

Thanks for giving me an opportunity to use it again.

A few parameters just have to be changed for each pattern and then

the program works out admissible values modulo each small prime.

> It looks like you have found: the first occurrence of the tightest

> possible bunching of 16 primes derived from the addititive

> combinations of 5 numbers.

Due to an initial mistake, I only found the first occurrence of

the bunching which is the tightest possible when the 4 smallest

numbers are all even. If some of them are allowed to be odd

(and I see no reason to disallow that) then I'm searching for

the first occurrence but may stop before succeeding.

I chose to search a specific tight bunching because I already

had a suitable program for a fixed pattern, and a tight bunching

is likely to have consecutive primes.

32 distinct primes generated by the additive combination of six

numbers seems relatively easy. There are large solutions based

on other published results. For example, any two non-overlapping

AP16 with the same common difference will give a solution.

http://hjem.get2net.dk/jka/math/aprecords.htm has more than needed:

AP19 and AP20 by Jaroslaw Wroblewski.

AP19: 254215977184797362303 + 53#*n, n = 0..18

AP20: 178284683588844176017 + 53#*n, n = 0..19

They have respectively 4 and 5 AP16 as subsequences so this

gives 4*5 = 20 AP16-based solutions in total.

The solution with the last AP16 in both cases gives 32 21-digit primes for

574731073635911261190, 21671067559381570778, 4*53#, 2*53#, 53#, 53#/2.

Here 574731073635911261190 +/- 21671067559381570778 is

in the middle of the two chosen AP16.

In 2004 Gennady Gusev and I published the first 3 AP17 with

common difference 17#:

http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0410&L=nmbrthry&F=&S=&P=2620
In total they give 2*2*3 = 12 pairs of AP16's.

By the way, the same tuple program as above was used.

--

Jens Kruse Andersen